为什么我们不需要在Fast Power算法中对每个操作数执行模运算？

Question

Today I practiced with a puzzle "fast power", which used a formula: (a * b) % p = (a % p * b % p) % p to calculate (a^n)%p , something like that: 2^31 % 3 = 2

However, I am so confused when I found the answer used ((temp * temp) % b * a) % b; to solved situation when n is odd, like 2^3

(temp is (temp * temp) % b * a recursively or (temp * temp) % b ).

Should not it be ((temp * temp) % b * a%b) % b ?

Since according to this formula, everything should %b before times together.

Answer 1

Should not it be ((temp * temp) % b * a % b) % b ?

No. For a , if you know beforehand that a won't overflow(a is smaller than b), you don't have to mod it.

The idea is modular arithmetic works for addition and multiplication. Operation like (a + b) % M = (a % M + b % M) % M and (a * b) % M = (a % M * b % M) % M are generally performed to avoid overflow of (a * b) and (a + b) and keep the value under certain range.

Example:

const int Mod = 7;
int a = 13;
int b = 12;
int b = b % Mod; // b now contains 5 which is certainly smaller than Mod

int x = (a % Mod * b) % Mod; // you won't need to mod b again if you know beforehand b is smaller than Mod

Update

C++ implementation of power function:

#define MOD 1000000007
// assuming x and n both be positive and initially smaller than Mod
int power(int x, int n) {
    if(n == 0) return x;
    int half = power(x, n / 2) % Mod;
    int ret = (half * half) % Mod; // you didn't need to do (half % Mod * half % Mod) % Mod because you already know half is smaller than Mod and won't overflow. 
                                   // Modulas being performed on the multiplied output, so now ret will be smaller than Mod
    if(n & 1) {
        ret = (ret * x) % Mod; // you didn't need to do (ret % Mod * x % Mod) % Mod
                               // because you already know ret and x is smaller than Mod
    }
    return ret;
}

Mod is an expensive operation. So you should avoid it whenever possible.